We call a compact complex manifold Moisezon manifold, if its dimension coincides with the algebraic dimension, i.e. it has as many algebraically independent meromorphical functions as its complex dimension. Boris Moisezon himself gave a proof to the following theorem: Let $X$ be a Moisezon mainfold, then for $X$ to be projective it is necessary and sufficient to be Kähler.

It was told that it could also be formulated as a criterion for projectivity:

A compact complex manifold is projective if and only if it is kähler and moisezon.

I didn't find any complete proof to the second formulation. Does someone know where to find it? Or at least how it works?

Edit: I already found two versions for the implication "projective=>moisezon" (necessity to be kähler is in fact clear), but i am also interested in alternatives, especially for the other one. On the one hand Huybrechts, Complex Geometry and on the other hand Wells, Moisezon spaces and the kodaira embeddingtheorem.

Next Edit: I don't want this to look like a jeaopardy question but, after postponing the problem I stumbled upon this:

There is a different proof by Thomas Peternell, given in "Algebraicity Criteria for Compact Complex Manifolds", Math. Ann. 275, 653-672 (1986). Theorem 1.4. states a slight variation of the theorem of Moisezon which is indeed equivalent. More precisely it states, that if there exist a real $(1,1)$-form $\omega$ and a real $2$-form $\varphi$ on a Moisezon manifold $X$ such that $\omega$ is positive definite, $d(\omega-\varphi) = 0$ and $\int_C \varphi = 0$ for all curves $C\subset X$, then $X$ is projective.


Are you asking why a projective complex variety has "enough" meromorphic functions to be Moisezon?

If so, the answer is that if $X$ is projective of dimension $d$, then the field of rational functions on $X$ (all of which are meromorphic --- and conversely, although we won't need this latter fact) is of transcendence dimension $d$, i.e. contains (exactly) $d$ algebraically independent elements. This is part of the basic dimension theory of algebraic varieties, and I'm pretty sure that it is discussed in Chapter I of Hartshorne's book (and in many other places too).

  • $\begingroup$ Of corse it is - how could i miss this. How to make yourself a fool....Thank you! $\endgroup$ – Ben Jul 21 '11 at 1:05
  • $\begingroup$ I shouldn't have accepted this, because an alternative for the even more interesting implication is still missing...I'm sorry for toggling $\endgroup$ – Ben Jul 21 '11 at 6:46
  • $\begingroup$ @Ben: Dear Ben, No problem. But which implication is "the even more interesting one"? (Sorry, I've just gotten myself confused about what it is you're trying to find out.) Best wishes, $\endgroup$ – Matt E Jul 21 '11 at 7:36
  • $\begingroup$ As you pointed, the "projective=>moisezon"-one is easy...Its the converse that i didn't find any proof of except the original (translated) by Moisezon himself...To prevent more confusion: I'm looking for another proof of "moisezon and kähler => projective". I thought of something like ~constructing a Hodge class... $\endgroup$ – Ben Jul 21 '11 at 8:26

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